What is the vertices 4x^2-8x+2y=5 ?

The vertices 4x^2-8x+2y=5 is Maximum (1, 9/2)

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vertices 4x^2-8x+2y=5

vertices

4 x^{2} - 8 x + 2 y = 5

Maximum (1, \frac{9}{2})

Solution steps

4 x^{2} - 8 x + 2 y = 5

Rewrite

4 x^{2} - 8 x + 2 y = 5

in the form

y = a x^{2} + b x + c

y = - 2 x^{2} + 4 x + \frac{5}{2}

The parabola parameters are:

a = - 2, b = 4, c = \frac{5}{2}

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 ( - 2 )}

Simplify

- \frac{4}{2 ( - 2 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = \frac{9}{2}

Therefore the parabola vertex is

(1, \frac{9}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (1, \frac{9}{2})

What is the vertices 4x^2-8x+2y=5 ?
The vertices 4x^2-8x+2y=5 is Maximum (1, 9/2)